The official solution looks like this, I think it's more clear. Z we can simply preprocess it (pay attention to the multiplication film), the test point for q is [fractional film] that is (a/b)%P = a* inverse of b %P
This involves calculating the inverse of b, and I use the Euclidean algorithm. The following blog post is very clear.
http://www.cnblogs.com/frog112111/archive/2012/08/19/2646012.html
Then there are two details. One is to write fast power so that Z can be preprocessed with O(k * log(n)) complexity. When fast power is used, note that the type of a must be long long, because when a*a It may explode, and I have been WA many times because of this. The second detail is that we need to judge the case of q=1 according to the summation formula of the proportional sequence
1 #include<iostream> 2 #define MAXN 200000 3 #define ll long long 4 #define P 1000000009 5 using namespace std; 6 7 int exgcd(int a,int b,int &x,int &y) 8 { 9 if(b==0) 10 { 11 x=1; 12 y=0; 13 return a; 14 } 15 int r=exgcd(b,a%b,x,y); 16 int t=x; 17 x=y; 18 y=t-a/b*y; 19 return r; 20 } 21 22 ll power(ll a,int n){ 23 if(n==0) return 1; 24 if(n==1) return a; 25 if(n==2) return (a*a)%P; 26 if(n%2) return power(power(a,n/2),2)*a%P; 27 return power(power(a,n/2),2); 28 } 29 30 31 int n,a,b,k,s[MAXN]; 32 int x,y; 33 ll sum; 34 35 int main(){ 36 37 scanf("%d%d%d%d",&n,&a,&b,&k); 38 39 for(int i=0;i<k;i++){ 40 char sym; cin>>sym; 41 if(sym=='+') s[i]=1; 42 else s[i]=-1; 43 } 44 45 for(int i=0;i<k;i++){//Z 46 if(s[i]==1) sum=(sum+power(a,n-i)*power(b,i)%P)%P; 47 else sum=(sum-power(a,n-i)*power(b,i)%P+P)%P; 48 } 49 50 exgcd(a,P,x,y);//找a的逆元 51 int inva=x; 52 inva+=P; inva%=P; 53 ll q = power(b,k)*power(inva,k)%P; 54 55 if(q==1){ 56 cout<<sum*((n+1)/k)%P; 57 } 58 else{ 59 x=0; y=0; 60 exgcd(q-1,P,x,y); x+=P; x%=P; 61 printf("%d",sum*x%P*( power(q,(n+1)/k) -1 )%P ); 62 } 63 64 return 0; 65 }